Question

An object of mass 1 kg is moving with a speed of 50 m/s. The object is brought to rest in 0.05 seconds by an external system. Find the change in momentum of the object and the average force applied by the external system.​

Answer 1

Given

Mass of the object = 1 kg.

Initial velocity of the object = u = 50 m /s

Final velocity of the object = v = 0 m/s.

To Find

1. Change in momentum.(∆p)
2. Average force applied

Calculations

initial momentum = mu

Final momentum = mv

Change in momentum = mv - mu

=> ∆p = m( v - u )

=> ∆p = 1( 0 - 50)

=> ∆p = -50 kg m/s.

Now, Ft = m(v-u) ( Since, impulse = change in momentum)

=> F = ∆p/t

=> F = 50/0.05

=> F = 1000 N .

Hence, change in momentum of the object is -50 kg m/s and force applied is 1000 N.

Answer 2

Given ,

Mass (m) = 1 kg

Initial velocity (u) = 50 m/s

Final velocity (v) = 0 m/s

Time (t) = 0.05 sec

We know that ,

The change in momentum is called impulse

$\boxed{ \sf{Impulse = m(v - u) }}$

Thus ,

Impulse = 1(0 - 50)

Impulse = -50 kg.m/s

Therefore ,

• The change in momentum of the object is 50 kg.m/s

Now , force is defined as the rate of change of momentum

$\boxed{ \sf{Force = \frac{Δp}{t} }}$

Thus ,

Force = 50/0.05

Force = 5000/5

Force = 1000 Newton

Therefore ,

• The average force applied by the external system is 1000 N